Exams › JEE Advanced › Maths

Define f, g: R -> R by f(x) = e^(x-1) - e^(-|x-1|) and g(x) = (1/2)(e^(x-1) + e^(1-x)). Find the area of the region in the first quadrant bounded by y = f(x), y = g(x) and x = 0.

1. (2 - sqrt(3)) + (1/2)(e - e⁻¹)
2. (2 + sqrt(3)) + (1/2)(e - e⁻¹)
3. (2 - sqrt(3)) + (1/2)(e + e⁻¹)
4. (2 + sqrt(3)) + (1/2)(e + e⁻¹)

Correct answer: (2 - sqrt(3)) + (1/2)(e - e⁻¹)

Solution

This standard JEE Advanced 2022 problem evaluates to (2 - sqrt3) + (1/2)(e - e⁻¹) after finding the intersection and integrating the difference of the curves over the bounded region.

Related JEE Advanced Maths questions

• What is the area enclosed by the curves y = 2 − |2 − x| and y = 3/|x|?
• If f(x) and g(x) are continuous functions over the interval a ≤ x ≤ b, and p(x) represents the greater of f(x) and g(x), while q(x) represents the lesser of f(x) and g(x), what is the expression for the area enclosed by the curves y = p(x), y = q(x), and the vertical lines x = a and x = b?
• The area enclosed by the curves y = sin x + cos x and y = |cos x − sin x| over the interval [0, π/2] is -
• Let f: [1/2, 1] → R (the set of all real numbers) be a positive, non-constant and differentiable function such that f'(x) < 2 f(x) and f(1/2) = 1. Then the value of ∫[1/2 to 1] f(x) dx lies in the interval -
• Let S = {(x, y) ∈ R × R: x ≥ 0, y ≥ 0, y² ≤ 4x, y² ≤ 12 - 2x and 3y + √8x ≤ 5√8}. If the area of the region S is α√2, then α is equal to -
• The area of the region enclosed by the curves y = |(x + 2)/(x - 2)| and y = |(x - 2)/(x + 2)| together with the x-axis equals 4 ln(a/e). Find the value of a.

⚔️ Practice JEE Advanced Maths free + battle 1v1 →